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2 years ago

Which equation represents the function graphed on the coordinate plane?

Mathematics

2 answers:

natita [175]2 years ago

5 0

Answer:

Step-by-step explanation:

The second option shows the function represented in the given graph.

Basically, the graph shows an absolute value function, because it has the form of a "V".

We know that the parent function of this is $f(x)=|x|$ , which is the simplest form of a value absolute function. So, we can get the right answer by just applying all given transformations to the parent function.

In the graph we see that the function was move 10 units downside, and 4 units to the left. This means that the function has to have "-10" and "+4", where the negative indicates the downside movement, and the positive sign indicates the left-side movement.

It's important to remember that, in case of horizontal displacement, with add to move to the left, and we subtract to move to the right.

Applying these translation to the parent function, we have

$g(x)=|x+4|-10$

Therefore, the right answer is option B.

Ulleksa [173]2 years ago

4 0

Answer:

The two lines connect at (-4,-10)This means you would add 4 to x and then subtract 10 from the equation.

This means the answer is: B.g(x) = |x + 4| – 10.

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Describe how to estimate a non-perfect square root to the hundredths place without using a calculator.

GaryK [48]

1) find the perfect square closest to the number
2) take that square root, and then divide your number by that number
3) take the average of the perfect square root (the whole number) and the result of dividing your original number by that root.

Example: Approximate square root of 27
Closest perfect square: 25
Square root: 5
27 ÷ 5 = 5.4
Average between 5 and 5.4 = 5.20
5.2² = 27.04. Fast and not a bad approximation!

8 0

2 years ago

Read 2 more answers

The life, in years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o

Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

$\bf P(X$

So, the probability that a switch fails in the first year is

$\bf P(X$

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and

$\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}$

where

$\bf \binom{100}{k}$ equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

$\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999$

3 0

2 years ago

Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.

Ede4ka [16]

The error is:

She incorrectly broke up 2 and two-fifths.

Option A is correct option.

Step-by-step explanation:

We need to describes Catron’s error?

The steps are:

Step 1: (Negative 9) (2 + two-fifths)

Step 2: (Negative 9) (Negative 2) + (negative 9) (two-fifths)

Step 3: Negative 18 + (negative 9) (two-fifths)

Step 4: (negative 27) (two-fifths)

Solution: Negative StartFraction 54 over 5 EndFraction = Negative 10 and four-fifths

The error is:

She incorrectly broke up 2 and two-fifths.

Option A is correct option.

<u>Solving </u>

Solving the equation by removing the error:

Step 1: (Negative 9) (2 + two-fifths)

taking LCM of 2 and two-fifths (2/5)

Step 2: (Negative 9) (twelve-fifths)

Step 3: (Negative 9*twelve)-fifths

Step 4: Negative one hundred and eight- fifths

Solution: Negative StartFraction -108 over 5 EndFraction = Negative one hundred and eight- fifths

Keywords: Solving Expressions

Learn more about Solving Expressions at:

#learnwithBrainly

5 0

2 years ago

Read 2 more answers

The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

zhannawk [14.2K]

Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

$(u+v)^2=u^2+2uv+v^2$ .

$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

$B=(18a+1)$

$C=(81-r^2)$

The discriminant is

$B^2-4AC$

$(18a+1)^2-4(a^2)(81-r^2)$

Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

$(u+v)^2=u^2+2uv+v^2$

$(324a^2+36a+1)-4a^2(81-r^2)$

Distribute the 4a^2 to the terms in the ( ) next to it:

$324a^2+36a+1-324a^2+4a^2r^2$

$36a+1+4a^2r^2$

We know that $a>0$ because the parabola is open up.

We know that $r>0$ because in order it to be a circle a radius has to exist.

So our discriminat is positive which means we have two solutions for $x^2$ .

But how many do we have for just $x$ .

We have to go further to see.

So the quadratic formula is:

$\frac{-B \pm \sqrt{B^2-4AC}}{2A}$

We already have $B^2-4AC}$

$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

To find $x$ we must square root both sides.

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

So there is only that one real solution (it actually includes 2 because of the plus or minus outside) here for x since the other one is square root of a negative number.

That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}$

$x=\pm \sqrt{\frac{0}{2a^2}}$

$x=\pm 0$

$x=0$

The equations intersect at x=0. Plugging into $y=ax^2+9$ we do get $y=a(0)^2+9=9$ .

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

$0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}$

A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

Add 18a+1 on both sides:

$18a+1=\sqrt{36a+1+4a^2r^2$

Square both sides:

$324a^2+36a+1=36a+1+4a^2r^2$

Subtract 36a and 1 on both sides:

$324a^2=4a^2r^2$

Divide both sides by $4a^2$ :

$81=r^2$

Square root both sides:

$9=r$

The radius is 9 as we stated earlier.

Let's go through the radius choices.

If the radius of the circle with center (0,0) is less than 9 then the circle wouldn't intersect the parabola. So It definitely couldn't be the last two choices.

7 0

2 years ago

Read 2 more answers

The graph shows the solution to the initial value problem y'(t)=mt, y(t0)= -4

Charra [1.4K]

$y'(t)=mt
\\
\\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C$

Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.

$y=at^2+b
\\
\\3=a\times 2^2-2
\\
\\3=4a-2
\\
\\5=4a
\\
\\a= \frac{5}{4} 
\\
\\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C
\\ \\ \frac{m}{2} = \frac{5}{4} 
\\
\\m= \frac{5}{2} 
\\
\\C=-2
\\
\\y(t_0)=-4
\\
\\-4= \frac{5}{4} t_0^2-2
\\
\\ -2= \frac{5}{4} t_0^2
\\
\\ t_0^2=- \frac{8}{5} 
\\
\\t_0=\pm \sqrt{- \frac{8}{5} }$

3 0

2 years ago